Answer
a) $x^2 \leq y^2$
b) $\sqrt{xy} \leq \frac{(x+y)}{2}$
Work Step by Step
a) Re-arrange the first inequality such as $x^2 \leq xy$ ...(1)
Also, $xy \leq y^2$ ....(2)
Now, let us combine the both inequalities (1) and (2).
Also, we have $x^2 \leq xy \leq y^2$
This gives: $x^2 \leq y^2$
b) Re-arrange the first inequality such as
Also, $\sqrt{xy} \gt \frac{(x+y)}{2}$
This gives: $xy \gt \frac{x^2+2xy+y^2}{4}$
or, $x^2-2xy+y^2 \lt 0$
$(x-y)^2 \lt 0$
As the square exprssion is alwsuys being non-negative.
Thus, we have $\sqrt{xy} \leq \frac{(x+y)}{2}$
